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(3p-2)(p+12)=0
We multiply parentheses ..
(+3p^2+36p-2p-24)=0
We get rid of parentheses
3p^2+36p-2p-24=0
We add all the numbers together, and all the variables
3p^2+34p-24=0
a = 3; b = 34; c = -24;
Δ = b2-4ac
Δ = 342-4·3·(-24)
Δ = 1444
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1444}=38$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(34)-38}{2*3}=\frac{-72}{6} =-12 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(34)+38}{2*3}=\frac{4}{6} =2/3 $
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